I recently read the book _The Man Who Loved Only Numbers_ by Paul Hoffman (the story of the prolific mathematician Paul Erdős). In his book, Hoffman mentions the Monty Hall problem and the fact that Erdős just couldn't believe the answer that Marilyn vos Savant gave in her weekly column in Parade magazine. Well, neither could I, and even after reading many web sites on the topic, including this one, I still believe she had it wrong.

Summary: We have 3 doors with a car behind one (the prize) and goats behind the other two (non-prizes). You pick one of the doors; Monty Hall then opens one of the remaining doors with a goat behind it (he never opens the door with the prize). The issue is: do you stick with your original choice or switch.

Marilyn says you should switch, as do many of the sites I've visited, claiming a 2/3 probability of win if you switch vs. 1/3 if you don't. The case analysis that Marilyn did is shown in Hoffman's book, and I believe it is flawed. She combined the two different action options of my Case A into a single option.

My decomposition of cases and their outcomes below shows that the probability of winning is the same (50%) whether you switch or not. The critical case is Case A in which the host (Monty) has two different doors he can open.

Here's my analysis:

(Note: I apologize for the clumsy table format, but having it in tabular form makes it much clearer. I tried html but couldn't get it to work correctly, and extra spaces are automatically deleted.)

I select Door #1, but I consider all cases of prize arrangement, so the same analysis applies for any initially selected door.

Option 1: Choose Door & Always Switch:
_____|______ Door ______| ___________Actions__________ | _______
Case |__1__|___2__|__3 _|_ 1stPick | HostOpens | 2ndPick _| Outcome
A ___| Car _| Goat_| Goat | ___ 1 __|_____ 2 ___|___ 3 ___|__ Lose
A ___| Car _| Goat_| Goat | ___ 1 __|_____ 3 ___|___ 2 ___|__ Lose
B ___| Goat |_ Car_| Goat | ___ 1 __|_____ 3 ___|___ 2 ___|__ Win
C ___| Goat | Goat_| Car _| ___ 1 __|_____ 2 ___|___ 3 ___|__ Win

Option 2: Choose Door & Always Stick with It:
_____|______ Door ______| ___________Actions__________ | _______
Case |__1__|___2__|__3 _|_ 1stPick | HostOpens | 2ndPick _| Outcome
A ___| Car _| Goat_| Goat | ___ 1 __|_____ 2 ___|___ 1 ___|__ Win
A ___| Car _| Goat_| Goat | ___ 1 __|_____ 3 ___|___ 1 ___|__ Win
B ___| Goat |_ Car_| Goat | ___ 1 __|_____ 3 ___|___ 1 ___|__ Lose
C ___| Goat | Goat_| Car _| ___ 1 __|_____ 2 ___|___ 1 ___|__ Lose

For both options, the probability of win is 1/2.

How is my analysis flawed?

Thanks,

Don Link
Columbia, MD